\(A=\left(-2a+3b-4c\right)-\left(-2a-3b-4c\right)\)

=\(-2a+3b-4c+2a+3b+4c\)

\(=\left(2a-2a\right)+\left(3b+3b\right)+\left(4c-4c\right)\)

\(=6b\)

Thay \(a=2012\) \(b=-1\)\(c=2013\)vào A ta đc

A=\(\left(-2.2012+3.-1-4.2013\right)-\left(-2.2012-3.-1-4.2013\right)\)

\(=-2012.2+-3-4.2013+2.2012-3+4.2013\)

=\(\left(2012.2-2.2012\right)+\left(-3-3\right)+\left(4.2013-4.2013\right)\)

\(=-6\)

Ta thấy \(\left\{\begin{matrix}\left(2x-1\right)^{2008}\ge0\\\left(y-\frac{2}{5}\right)^{2008}\\\left|x+y-z\right|\ge0\end{matrix}\right.\ge0\)

\(\Rightarrow\left(2x-1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y-z\right|\ge0\)

Mà theo đề ra

\(\left(2x-1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y-z\right|=0\)

\(\Rightarrow\left\{\begin{matrix}2x-1=0\\y-\frac{2}{5}=0\\x+y-z=0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}2x=1\\y=\frac{2}{5}\\z=x+y\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=\frac{1}{2}\\y=\frac{2}{5}\\z=\frac{2}{5}+\frac{1}{2}=\frac{9}{10}\end{matrix}\right.\)

Vậy \(x=\frac{1}{2}\) y=\(\frac{2}{5}\)và z=\(\frac{9}{10}\)

cầm đầu